#### Question

The temperature of 170g of water at 50°C is lowered to 5°C by adding a certain amount of ice to it. Find the mass of ice added.

#### Solution

Given data:

M_w = 170 gm,S_w = 1 `"cal/gm"^@C` = 4200 J / `kg^@C`

`triangle T = 5^@C, L_f = 80 cal/gm` = 336000 J/kg

Amount of heat given by water = Amount of heat absorb by ice .....[Principle of calorimetry]

`M_wS_wtriangleT = M_"ice" xx L_f`

`(170)(1)(5) = M_"ice" xx (80)`

:. `M_"ice" = (170xx1xx5)/(80)` = 10.625 gm

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#### APPEARS IN

Solution The Temperature of 170g of Water at 50°C is Lowered to 5°C by Adding a Certain Amount of Ice to It. Find the Mass of Ice Added. Concept: Calorimetry - Specific Heat Capacity.